A particular type of gasoline is supposed to have a mean oct

A particular type of gasoline is supposed to have a mean octane rating of 90%. Five measurements are taken of the octane rating and the results (in %) are 87.0, 86.0, 86.5, 88.0, 85.3.

Test the requirements of the octane rating using the appropriate hypothesis test (assume alpha=0.01).

Verify your result in part (a) using the appropriate confidence interval.  

Solution

sample size n=5

sample mean=86.56

sample standard deviation =1.021274

Let mu be the population mean

The test hypothesis:

Ho: mu=90 (i.e null hypothesis)

Ha: mu not equal to 90 (i.e. alternative hypothesis)

The test statistic is

t=(xbar-mu)/(s/vn)

=(86.56-90)/(1.021274/sqrt(5))

=-7.53

The degree of freedom =n-1=5-1=4

It is a two-tailed test.

Given a=0.01, the critical values are t(0.005, df=4) =4.6 or -4.6 (from student t table)

The rejection regions are if t<-4.6 or t>4.6, we reject the null hypothesis.

Since t=-7.53 is less than -4.6, we reject the null hypothesis.

So we can not conclude that A particular type of gasoline is supposed to have a mean octane rating of 90%

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So the lower bound is

xbar - t*s/vn = 86.56 - 4.6*(1.021274/sqrt(5)) = 84.45905

So the upper bound is

xbar + t*s/vn =86.56 + 4.6*(1.021274/sqrt(5))=88.66095

Since all values on the interval are less than 90, we can not conclude that A particular type of gasoline is supposed to have a mean octane rating of 90%